Question

我试图在一个请求中得到两个请求的结果，这两个后续请求是有效的，并且每个请求都产生一个包含两列的表：

SELECT patron.last_name, COUNT(*) AS **pret**
FROM circ_transaction_log INNER JOIN patron ON     circ_transaction_log.patron_id=patron.patron_id
AND **circ_transaction_log.transaction_type<5**
AND patron.college_or_school = 'High School'
GROUP BY patron.last_name;

last_name |的 pret的
________________
史蒂文格雷勒| 552个
米歇尔葡萄酒| 122 ...

_________________________ 或 _________________________

SELECT patron.last_name, COUNT(*) AS **resa**
FROM circ_transaction_log INNER JOIN patron ON    circ_transaction_log.patron_id=patron.patron_id
AND **circ_transaction_log.transaction_type BETWEEN 5 AND 10**
AND patron.college_or_school = 'High School'
GROUP BY patron.last_name;

last_name |的 RESA
________________
史蒂文格雷勒| 12个
米歇尔葡萄酒| 8
...

我想得到的结果有点像这样：

last_name | resa |的 pret的
_______________________________
史蒂文格雷勒| 552 | 12个
米歇尔葡萄酒| 122 | 8
......

但我认为困难在于我要求同一张桌子两次（表CIRC_TRANSACTION_LOG）以及我尝试的任何内容或错误或不起作用

预先感谢您的回复

此致 Nickk

Answer 1

你是这样的，然后：

SELECT   p.last_name, 
         COUNT(case when ctl.transaction_type < 5 then 1 end) AS pret,
         count(case when ctl.transaction_type between 5 and 10 then 1 end) as resa
FROM     circ_transaction_log ctl
         INNER JOIN patron p ON (ctl.patron_id = p.patron_id)
AND      ctl.transaction_type <= 10  -- possibly not required if transaction_type is always <= 10 
AND      p.college_or_school = 'High School'
GROUP BY p.last_name;

NB。未经测试，因为您没有为表格提供任何样本数据。

Answer 2

我更喜欢这种方法。如果该行符合您的条件，请在列中添加1，如果它不匹配，则将0添加。然后，当您将所有值相加时，它会为您提供匹配的行数。这允许您为不同的条件创建不同的总和。

SELECT
    PatronCounts.last_name,
    SUM(PatronCounts.Pret) Pret,
    SUM(PatronCounts.Resa) Resa
FROM
(SELECT 
    patron.last_name, 
    CASE WHEN circ_transaction_log.transaction_type < 5 THEN 1 ELSE 0 END AS Pret,
    CASE WHEN circ_transaction_log.transaction_type BETWEEN 5 AND 10 THEN 1 ELSE 0 END AS Resa,
FROM 
    circ_transaction_log 
    INNER JOIN patron ON circ_transaction_log.patron_id=patron.patron_id 
WHERE
    patron.college_or_school = 'High School'
GROUP BY 
    patron.last_name) PatronCounts

Answer 3

您可以使用（last_name）对结果表的完全外部联接组合您的查询。这应该产生预期的结果。试图创建查询...

Select * from (SELECT patron.last_name, COUNT(*) AS pret FROM circ_transaction_log INNER JOIN patron ON circ_transaction_log.patron_id=patron.patron_id AND circ_transaction_log.transaction_type<5 AND patron.college_or_school = 'High School' GROUP BY patron.last_name) FULL OUTER JOIN (SELECT patron.last_name, COUNT(*) AS resa FROM circ_transaction_log INNER JOIN patron ON circ_transaction_log.patron_id=patron.patron_id AND circ_transaction_log.transaction_type BETWEEN 5 AND 10 AND patron.college_or_school = 'High School' GROUP BY patron.last_name) USING (last_name)

仅通过一个请求替换2个SQL / ORACLE请求

3 个答案: